Math Doubts

Evaluate $\displaystyle \Large \lim_{x \,\to\, \pi} \, \normalsize \dfrac{1-\cos{7(x-\pi)}}{5{(x-\pi)}^2}$

$x$ is a literal number but considered it to denote the angle of a right angled triangle. The quotient of subtraction of $\cos{7(x-\pi)}$ from one by the product of $5$ and ${(x-\pi)}^2$ is expressed in mathematical form as follows.

$\dfrac{1-\cos{7(x-\pi)}}{5{(x-\pi)}^2}$

The limit of the function is written in the following form as $x$ tends to pi.

$\displaystyle \Large \lim_{x \,\to\, \pi} \, \normalsize \dfrac{1-\cos{7(x-\pi)}}{5{(x-\pi)}^2}$

01

Check the function

Substitute $x$ is equal to $\pi$ to study the functionality of the algebraic trigonometric function as $x$ tends to $\pi$.

$=\,\,\,$ $\dfrac{1-\cos{7(\pi-\pi)}}{5{(\pi-\pi)}^2}$

$=\,\,\,$ $\dfrac{1-\cos{7(0)}}{5{(0)}^2}$

$=\,\,\,$ $\dfrac{1-\cos{0}}{5 \times 0}$

$=\,\,\,$ $\dfrac{1-1}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

The value of the function is indeterminate when the value of the $x$ approaches $\pi$.

02

Transforming the function

Take $u = x-\pi$.

If $x \to \pi$, then $x-\pi \to 0$. Therefore, $u \to 0$. Now, transform the entire mathematical expression in terms of $u$ purely.

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{1-\cos{7u}}{5u^2}$

03

Simplifying the function

Transform the numerator to apply subtraction trigonometric rule of subtraction of cos of double angle from one.

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{1-\cos{\Bigg(\dfrac{2}{2} \times 7u\Bigg)}}{5u^2}$

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{1-\cos{2 \Bigg(\dfrac{7u}{2}\Bigg)}}{5u^2}$

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{2\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{5u^2}$

$=\,\,\,$ $\displaystyle \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{2}{5} \times \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{u^2}$

Make some adjustments to denominator to be appeared same as the angle of the sine function in numerator.

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(u \times \dfrac{7}{2} \times \dfrac{2}{7}\Bigg)}^2}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2} \times \dfrac{2}{7}\Bigg)}^2}$

Use the product rule of exponents to expand the denominator.

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2 \times {\Bigg(\dfrac{2}{7}\Bigg)}^2}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2 \times \dfrac{4}{49}}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2} \times \dfrac{1}{\dfrac{4}{49}}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2} \times \dfrac{49}{4}$

$=\,\,\,$ $\displaystyle \dfrac{2}{5} \times \dfrac{49}{4} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\displaystyle \dfrac{2 \times 49}{5 \times 4} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\require{cancel} \displaystyle \dfrac{\cancel{2} \times 49}{5 \times \cancel{4}} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\require{cancel} \displaystyle \dfrac{49}{5 \times 2} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

$=\,\,\,$ $\require{cancel} \displaystyle \dfrac{49}{10} \times \Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin^2 {\Bigg(\dfrac{7u}{2}\Bigg)}}{{\Bigg(\dfrac{7u}{2}\Bigg)}^2}$

Apply, the quotient rule of exponents to simplify it further.

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times \Large \lim_{u \,\to\, 0} \, \normalsize {\Bigg(\dfrac{\sin {\Big(\dfrac{7u}{2}\Big)}}{\Big(\dfrac{7u}{2}\Big)} \Bigg)}^2$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times {\Bigg(\Large \lim_{u \,\to\, 0} \, \normalsize \dfrac{\sin {\Big(\dfrac{7u}{2}\Big)}}{\Big(\dfrac{7u}{2}\Big)} \Bigg)}^2$

04

Obtaining required result

Try limit of $\dfrac{\sin x}{x}$ rule as $x$ tends to $0$ to get the required result. In order to apply this formula, transform our expression to this form.

If $u \to 0$, then $\dfrac{7}{2} \times u \to \dfrac{7}{2} \times 0$. Therefore, $\dfrac{7u}{2} \to 0$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times {\Bigg(\Large \lim_{\huge \frac{7u}{2} \Large \,\to\, 0} \, \normalsize \dfrac{\sin {\Big(\dfrac{7u}{2}\Big)}}{\Big(\dfrac{7u}{2}\Big)} \Bigg)}^2$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times {(1)}^2$

$=\,\,\,$ $\displaystyle \dfrac{49}{10} \times 1$

$=\,\,\,$ $\displaystyle \dfrac{49}{10}$

$=\,\,\, 4.9$

It is the required result for this limit problem of the calculus mathematics.

Latest Math Topics
Latest Math Problems
Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved